# How many grams are there in #7.50 x 10^23# molecules of #H_2SO_4#?

Well, how many grams in

Thus, we calculate the quotient.

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To find the mass of (7.50 \times 10^{23}) molecules of (H_2SO_4), you first calculate the molar mass of (H_2SO_4) and then use Avogadro's number to convert the number of molecules to moles, and finally use the molar mass to find the mass in grams.

Molar mass of (H_2SO_4): (2 \times 1.008 , \text{g/mol (for H)} + 32.06 , \text{g/mol (for S)} + 4 \times 16.00 , \text{g/mol (for O)} = 98.08 , \text{g/mol})

Number of moles of (H_2SO_4): (\frac{7.50 \times 10^{23} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mol}} = 1.246 , \text{mol})

Mass of (H_2SO_4): (1.246 , \text{mol} \times 98.08 , \text{g/mol} = 122.26 , \text{g})

So, there are (122.26 , \text{grams}) in (7.50 \times 10^{23}) molecules of (H_2SO_4).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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